The volume of a cylinder is calculated by:

$\displaystyle V_{cyl}=\pi \cdot r^2 \cdot h$

With your problem you have r = 10, h = 10. That means the maximum volume is:

$\displaystyle V_{max}= \pi \cdot (10\ m)^2 \cdot (10\ m)=1000 \pi \ m^3$

The actual volume is three quarter of the maxium volume:

$\displaystyle V_{actual} = \frac34 \cdot V_{max} = 750 \pi \ m^3$ that means the surface of the oil is 7.5 m above ground. So the height above the leak is 3 m:

$\displaystyle h(0)= 3\ m$

$\displaystyle h(1)=3 \cdot 0.9 \ m$

$\displaystyle h(2)=(3 \cdot 0.9 \ m)\cdot 0.9 = 3 \cdot 0.9^2 \ m$

...

$\displaystyle h(t)=3 \cdot 0.9^t \ m$

2. Calculate the volume which has already leaked from the tank:

$\displaystyle V_{leak}=\pi \cdot 10^2 \cdot (3 - h(t)) = \pi \cdot 10^2 \cdot (3 - 3 \cdot 0.9^t) $

Since 5000 L = 5 m³ you have to solve for t:

$\displaystyle 5 = \pi \cdot 10^2 \cdot (3 - 3 \cdot 0.9^t)$

$\displaystyle 5=300\pi - 300\pi \cdot 0.9^t$

$\displaystyle \frac{5-300 \pi}{-300 \pi} = 0.9^t$

Use logarithms to calculate t. I've got t ≈ 0.0505 h and

**that are nearly 3 min** (This result seems a little funny to me

I assume that I've made a silly mistake with my calculations. So check!