Hello Temitope,
Let's begin with the formula for the volume of a pyramid:
$$V=\frac{1}{3}bh$$
Now, the base $b$ is a square, and so let's let $s$ be the measure of the sides of the square, and so our volume formula becomes:
$$V=\frac{1}{3}s^2h$$
At any point in time, the volume of water will be a square pyramid that is similar to the container, and so we know we will always have:
$$\frac{s}{h}=\frac{13}{10}\implies s=\frac{13}{10}h$$
Now we may express the volume in terms of one variable $h$:
$$V=\frac{1}{3}\left(\frac{13}{10}h\right)^2h=\frac{169}{300}h^3$$
Now, if we differentiate with respect to time $t$, we obtain:
$$\frac{dV}{dt}=\frac{169}{100}h^2\frac{dh}{dt}$$
Now, we are told:
$$\frac{dV}{dt}=10\frac{\text{m}^3}{\text{s}}$$
And so (given that our units of length are in meters and our units time in seconds) we may write:
$$10=\frac{169}{100}h^2\frac{dh}{dt}$$
Solving for $$\frac{dh}{dt}$$, we obtain:
$$\frac{dh}{dt}=\frac{1000}{169h^2}$$
And so we find that when the pyramid is full, or when $h=10$, we have:
$$\bbox[7px,border:2px solid #207498]{\left.\frac{dh}{dt}\right|_{h=10}=\frac{1000}{169(10)^2}\frac{\text{m}}{\text{s}}=\frac{10}{169}\frac{\text{m}}{\text{s}}}$$
