Essential Bandwidth of rect(t/T)
I'll add another answer, even though MBaz's answer is correct, because I think that it doesn't actually address the problem you have with arriving at the final solution. Summarizing, you have to solve

$$\frac{1}{2\pi}\int_{-W}^W|P(\omega)|^2d\omega=0.9\int_{-\infty}^{\infty}\text{rect}^2(t/T)dt=0.9\cdot T\tag{1}$$

for $W$ (which is the bandwidth in radians). If you like, rewrite (1) using $f$ instead of $\omega$. $P(\omega)$ is the Fourier transform of the rectangular pulse:

$$P(\omega)=2\frac{\sin(\omega T/2)}{\omega}\tag{2}$$

If I understood your question and your comments correctly, then you understand all of this, but your problem is solving (1) for $W$. Note that the integral has no closed-form solution, unless you consider an expression including the sine integral $\text{Si}(x)=\int_0^x\sin(t)/t\;dt$ closed form. With a little help from WolframAlpha, Equation (1) can be written as

$$\frac{2}{\pi W}\left[TW\cdot \text{Si}(TW)+\cos(TW)-1\right]=0.9\cdot T\tag{3}$$

Equation (3) has no closed-form solution for $W$, so it is not surprising that you got confused. You could of course try to solve (3) numerically but I do not think that this is the idea of the exercise. What you need to know is that obviously the largest portion of the energy is contained in the main lobe of the spectrum, i.e. between $-W_0$ and $W_0$, where $W_0$ is the first zero crossing of $P(\omega)$. From (2) it is clear that

$$W_0T/2=\pi\quad\Longrightarrow W_0=\frac{2\pi}{T}\tag{4}$$

So this is the essential bandwidth in radians.
Checking the exact percentage using the left-hand side of (3) we get

$$\frac{2}{\pi W_0}\left[TW_0\cdot \text{Si}(TW_0)+\cos(TW_0)-1\right]=
\frac{2\;\text{Si}(2\pi)}{\pi}\cdot T=0.9028\cdot T$$

which seems close enough.

In sum, the essential bandwidth of a rectangular pulse is given by the width of  the mainlobe of its spectrum, so you only need to be able to calculate the first zero of the spectrum and you're done.