Answer
$ t=\dfrac{ R^2 }{ r^2 } \sqrt{\dfrac{2d}{ g} } $
Work Step by Step
We can use Bernoulli’s principle since we know that a streamline connects every point on the surface (point 1) of the liquid to a point in the drain (point 2).
$$\color{red}{\bf\not}P_1+\frac{1}{2}\rho v_1^2+\rho gy_1=\color{red}{\bf\not}P_2+\frac{1}{2}\rho v_2^2+\rho gy_2$$
It is obvious that $P_1=P_2=P_a$ since both points are exposed to air.
Also, from the figure below, $y_1=y$ where $y$ changes with time, and $y_2=0 $.
$$ \frac{1}{2}\color{red}{\bf\not}\rho v_1^2+\color{red}{\bf\not}\rho gy = \frac{1}{2}\color{red}{\bf\not}\rho v_2^2+0\tag{$\times 2$}$$
$$v_2^2=v_1^2+2gy\tag 1$$
Now we can use the continuity equation since it relates the flow rates at the surface and at the hole.
$$Q=A_1v_1=A_2v_2$$
Noting that $A_1=\pi R^2$ and $A_2=\pi r^2$
Hence,
$$\pi R^2v_1=\pi r^2v_2$$
So,
$$v_1=\dfrac{v_2r^2}{R^2}$$
Plugging into (1);
$$v_2^2=\left[\dfrac{v_2r^2}{R^2}\right]^2+2gy $$
$$v_2^2= \dfrac{v_2^2r^4}{R^4}+2gy $$
$$v_2^2- \dfrac{v_2^2r^4}{R^4}=2gy $$
$$v^2\left[1-\left(\dfrac{ r }{R }\right)^4\right] =2gy $$
And we know that $r\lt \lt R$. Hence, $\left(\dfrac{ r }{R }\right)^4\rightarrow0$
Therefore,
$$v_2=\sqrt{2gy}\tag 2$$
Now we can find the volume flow rate at the whole which is given by
$$Q=\dfrac{dV}{dt} =A_2v_2$$
$$\dfrac{dV}{dt} =\pi r^2 v_2$$
So,
$$dt=\dfrac{dV}{\pi r^2 v_2}$$
Plugging from (2);
$$dt=\dfrac{dV}{\pi r^2 \sqrt{2gy } }$$
Noting that the volume of the water decreases with time due to the leak at the bottom hole. So, $dV=-A_1 dy=$
$$dt=\dfrac{-\pi R^2dy}{\pi r^2 \sqrt{2gy } }$$
$$dt=\dfrac{- R^2 }{ r^2 \sqrt{2gy } }dy$$
$$dt=\dfrac{- R^2 }{ r^2 \sqrt{2g } }\left[y^{-\frac{1}{2}}\right] dy$$
Integrating both sides;
$$\int_0^t dt=\int_{d}^0\dfrac{- R^2 }{ r^2 \sqrt{2g } }\left[y^{-\frac{1}{2}}\right] dy $$
$$ t=\dfrac{- R^2 }{ r^2 \sqrt{2g } }\int_{d}^0\left[y^{-\frac{1}{2}}\right] dy $$
$$ t=\dfrac{- R^2 }{ r^2 \sqrt{2g } } \left[2y^{ \frac{1}{2}}\right] \bigg|_{d}^0 $$
$$ t=\dfrac{- 2R^2 }{ r^2 \sqrt{2g } } \left[0-\sqrt{d}\right] $$
$$ t=\dfrac{ 2R^2 }{ r^2 } \sqrt{\dfrac{d}{2g} } $$
$$\boxed{ t=\dfrac{ R^2 }{ r^2 } \sqrt{\dfrac{2d}{ g} } } $$
