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GOLDEN GIRL 80 COMPRESSION 2-PIECE SURLYN COVER GOLF BALLS 15 GOLF BALL SET NEW    US $39.99

3 BRAND NEW PGA GOLF BALLS SURLYN COVER WHITE TRUCKS    US $8.00

3 NEW PGA GOLF BALLS SURLYN COVER WHITE TRUCKS LOGO    US $8.00

MAXFLI REVOLUTION GOLF BALLS - ONE DOZEN WITH TWO PIECE IONOMER COVER - NEW    US $14.95

Lot Of 13 Intech Beta TI Thin cover Pink Golf Balls    US $17.00

NIB Three Golf Balls Titleist X-Outs NEW Cover Surlyn    US $3.49

Top Flite D 2 Distance Firmer Cover Golf Balls 3 pk 9 balls    US $12.98

12 - Top Flite D2 Distance Firmer Cover Golf Balls New    US $19.97

TITLEIST X-0UTS MFG. BY TITLEIST GOLF DIVISION 3 GOLF BALLS W/COVER OF SURILYN    US $3.99

Srixon Distance X Golf Balls Soft Ionomer Cover 432 Dimples White One Dozen New    US $19.99

Titleist DT Solo Soft Long Distance Surlyn Cover Golf Balls One Dozen New    US $19.99

3 BRAND NEW PGA GOLF BALLS SURLYN COVER W/ LOGO    US $8.00

(3) Vintage Unopened Yellow 2 Cadwell Cover Tom Salvate Golf Balls    US $7.99

(3) Vintage Unopened Yellow 1 Cadwell Cover Tom Salvate Golf Balls    US $7.99

HUGE LOT NEW GOLF BALLS TITLEIST NIKE WILSON PRO TOP FLITE TOUR TITANIUM COVER    US $31.00

36 Thin Cover Intech Beta TI Ultra High Velocity Golf Balls NIP    US $12.99

Precept MC Double Cover Golf Balls-1 doz.    US $20.00

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Clair H gives the land area of Canada as 9,970,610 km².
1 km² = 10^12 mm², so
9,970,610 km² = 9.970610*10^18 mm²
Let's assume that this area is contiguous and made up of many small equilateral triangles which can be rearranged to comprise one giant equilateral triangle. The area of an equilateral triangle is
A = ((√3)/4)s^2
s = 2√((A)/√3)
so the side of our giant triangle would be
s = 2√((9.970610*10^18 mm²)/√3)
s ≈ 2,399,277,884.5904395457 mm
Dividing the side length by the diameter of the regulation golf ball, 42.67 mm, yields a side length of
n ≈ 56,228,682 balls
To make sure no golf balls fall off the land we make this our base number. (We could use (n + 1) if we assumed the center of a ball could perch on the edge of the land.) We also assume a flat Earth and no side rails, so our balls will form a truncated tetrahedron.
The bottom layer will have
n(n + 1)/2 , or 3,161,664,735,685,806 balls
Layer 2: 1,580,832,311,614,221 balls
Unfortunately, we reach a complete tetrahedron of
2.9629374556136784809884e+22 balls before we reach Avogadro's Number, N ≈ 6.02214179*10^23, so side rails are in order, after all. Using side rails we can alternate between layer 1 and 2 until we reach N.
(602214179000000000000000) / 4742497047300027 ≈ 126,982,510.056 or 253,965,020.11228 layers. Rounding up, we have 253,965,021 layers required. If we don't consider close-pack, this would be 190,473,763 layers times 46.77 mm ≈ 8,908,457,895.51 mm ≈ 8,908 km. Considering close-pack, h = (253,965,020)(0.816496581)(46.77) + 46.77 ≈ 9,698,300,746 mm ≈ 9,698 km, which is deeper than the stacking solution (lots of waste space in the alternate layers.

This leads to "What happens if we use a square base?".
√9.970610*10^18 mm² ≈ 3,157,627,273.7611 mm
(3,157,627,273.7611 mm) / (46.77 mm) ≈ 67,513,946 balls per side
67,513,946^2 = 4,558,132,904,490,916 balls per layer
(6.02214179*10^23) / 4,558,132,904,490,916 ≈ 132,118,609.00 layers
(132,118,609.00 layers)(46.77 mm) ≈ 6,179,187,342.93 mm ≈ 6,179.187 km

This is by far the smallest height.