The rate of water flow through a hole is given by Torricelli's theorem: Q = A × v, where A is the area of the hole and v is the speed of efflux, $\mathrm{v}=\sqrt{2\mathrm{gh}}$.

The flow rates from the two holes are equal.

For the square hole: Area A_sq = L², depth h_sq = 4y.

So, ${\mathrm{Q}}_{\mathrm{sq}}={\mathrm{L}}^2\ast \sqrt{2\mathrm{g}\left(4\mathrm{y}\right)}={\mathrm{L}}^2\ast \sqrt{8\mathrm{gy}}$

For the circular hole: Area A_cir = πR², depth h_cir = y.

So, ${\mathrm{Q}}_{\text{cir }}={\mathrm{\pi R}}^2\times \sqrt{2\mathrm{gy}}$

Set them equal: ${\mathrm{L}}^2\sqrt{8\mathrm{gy}}={\mathrm{\pi R}}^2\sqrt{2\mathrm{gy}}$

Simplify: ${\mathrm{L}}^2\sqrt{4}={\mathrm{\pi R}}^2$

 so 2L² = πR². Therefore,

$\text{R=}\frac{\mathrm{L}}{\sqrt{\frac{\mathrm{\pi }}{2}}}=\frac{\mathrm{L}}{\sqrt{2\mathrm{\pi }}}$

Final Answer: $\frac{\mathrm{L}}{\sqrt{2\mathrm{\pi }}}$