It's not a diode. Just doped semiconductor (silicon.)
I'll be using the problem value for \$q\$. Yours is more precise. But I want to keep it simpler, for now.
getting an answer
The value of \$n_i=1\times 10^{10}\:\frac{\text{e}^-}{\text{cc}}\$ is typical for pure intrinsic silicon at \$T=300\:\text{K}\$. (GaAs has \$E_g\approx 1.4\:\text{eV}\$ and so would be much, much fewer conduction band electrons, for example.) So I conclude this is a silicon cube.
Without getting into the whys and wherefores about the mass-action law, you know that \$n\:p=n_i^2\$, given that pretty much all donors/acceptors are all ionized at \$T=300\:\text{K}\$, given that boron is an acceptor, and given that \$\left(p=1\times 10^{15}\right)\gg \left(n_i=1\times 10^{10}\right)\$, you can find that \$n=\frac{n_i^2}{p}=1\times 10^5\$.
Since \$\left(p=1\times 10^{15}\right)\gg \left(n=1\times 10^{5}\right)\$, ignore \$n\$ and just focus on \$p\$. So just use \$\mu_p\$ and ignore \$\mu_n\$: \$\sigma=q\,p\,\mu_p= 0.08 \:\frac{\mho}{\text{cm}}\$. That, times the applied volts, times the cross-section area, divided by the length gives you the current: \$800\:\mu\text{A}\$.
(The part associated with \$q\,n\,\mu_n\$ is so small as to be ignorable -- about \$160\:\text{fA}\$.)
looking over your work
When you tried to compute \$\sigma\$ you were, instead, computing the current density, \$J=\sigma\,\mathscr{E}\$.
You are supposed to use \$p\$ and not \$n_i\$. Another mistake.
You also screwed up on the units. It is true that \$\mu_p=500\:\frac{\text{cm}^2}{\text{V}\,\cdot\,\text{s}}\$. But note the use of \$\text{cm}\$. Meanwhile, you applied \$\mathscr{E}=1\times 10^6\:\frac{\text{V}}{\text{m}}\$. But that's using \$\text{m}\$ and not \$\text{cm}\$. So you mixed up your units, even for computing \$J\$.
So even if you knew you were computing \$J\$ (which you don't appear to have known), you still got it wrong.
\$J=800\:\frac{\text{A}}{\text{cm}^2}\$.
Using \$J\$ instead, just to take a different tack, then \$I=A\,J=\left(10\:\mu\text{m}\right)^2\cdot 800\:\frac{\text{A}}{\text{cm}^2}=800\:\mu\text{A}\$.
summary
I think the answer should be \$I=800\:\mu\text{A}\$. (Give or take a little.)