Let the capacity of tank be $1$ litre.
Draining rate $=\dfrac{0.5\text{ litre}}{30\text{ minutes}}=\dfrac{1}{60} \text{ litre/min}$
Let filling rate be $x \text{ litre/min}$
In $1$ min tank gets $x- \left(\dfrac{1}{60}\right) \text{litre}$ filled.
To fill the remaining half part we need $10 \text{ minutes}$
$x-\dfrac{1}{60} \text{ litre}\to 1 \text{ min}$
$0.5 \text{ litre}\to 10 \text{ mins}$
$\frac{0.5}{\left(x-\frac{1}{60}\right)}=10$
Solving, we get $x= \dfrac{4}{60}$ which is $4$ times more than draining rate.
So, option A