The power absorbed by the BOX in the Fig.A is $p(t) = 2.5e^\left(-4t\right)$ W. Compute the energy and charge delivered to the BOX in the time interval $0 < t < 250$ ms.
So let me show you my try:
A. Find the power absorbed by finding the integral of $p(t) = 2.5e^\left(-4t\right)$ from 0 to 250 $ms$ that is 0.25 $s$
$$\begin{align} \int_0^\left(0.25\right) p(t) = 2.5e^\left(-4t\right) dt \end{align}$$
$p=0.395$ W
B. Find the energy delivered to the BOX
$E=p\times t$
$E=0.395 \times 0.25 = 0.009J = 9$ $mJ$
C. Find the charge delivered to the BOX
$V=50e^\left(-t\right)=38.94$ $V$
$i=\frac pV=\frac {0.395} {38.94}=0.01$ $A$
$Q=i \times t= 0.01 \times 0.25=0.0025$ $C= 2.5$ $mC$
The correct answer in the book was:
395.1 $mJ$, 8.8 $mC$.
There seem to be many errors in your attempt. Some of them could be typos, but there seem to be many conceptual issues as well (multiplying instead of integrating, etc.).
The energy absorbed by the box is the integral of the power with respect to time with the bounds being given by the time interval.
So the answer to part a) (energy absorbed) is simply:
$$E = \int_0^{0.25} 2.5e^{-4t}\ dt = -\frac{1}{4}(2.5)(e^{-1} - 1) \approx 0.3951J = 395.1mJ $$
For the answer to the second part, first find an expression for the instantaneous current $i(t)$. The relationship between instantaneous power, voltage and current is $P(t) = V(t)i(t)$.
So $\displaystyle i(t) = \frac{2.5e^{-4t}}{50e^{-t}} = 0.05e^{-3t}$
Now the charge delivered over that interval, $C$ is given by:
$$C = \int_0^{0.25} i(t)\ dt = \int_0^{0.25} 0.05e^{-3t} \ dt = -\frac{1}{3}(0.05)(e^{-0.75} - 1) \approx 0.0088C = 8.8mC$$