1. Use continuity of flow

At any instant, the volume flow rate leaving the tank through the tap must equal the rate at which the water level in the tank falls:

$A\left(-\frac{dh}{dt}\right)=av$

where:

• $A = 750\,\text{cm}^2$ = cross-sectional area of tank
• $a = 500\,\text{mm}^2$ = cross-sectional area of tap
• $v = 30\,\text{cm/s}$ = speed of water through the tap

So,

$\frac{dh}{dt}=-\frac{av}{A}$

2. Convert units carefully

Tap area:

$500\,\text{mm}^2 = 500\times 0.01\,\text{cm}^2 = 5\,\text{cm}^2$

since $1\,\text{cm}^2 = 100\,\text{mm}^2$.

Now substitute:

$\frac{dh}{dt}=-\frac{5\times 30}{750}\,\text{cm/s}$

$\frac{dh}{dt}=-\frac{150}{750}\,\text{cm/s}=-0.2\,\text{cm/s}$

3. Convert to m/s

$-0.2\,\text{cm/s}=-0.2\times 10^{-2}\,\text{m/s}=-2\times 10^{-3}\,\text{m/s}$

Thus,

$\frac{dh}{dt}=-2\times 10^{-3}\,\text{m/s}$

Comparing with

$\frac{dh}{dt}=x\times 10^{-3}\,\text{m/s}$

we get

$x=-2$

4. About the sign

Since the water level is falling, $\frac{dh}{dt}$ must be negative. Hence the physically correct value is $x=-2$.

If the question expects only the magnitude, then it would be $2$.