= Solution
Assume one collected electron per absorbed photon, ideality factor one and $T=298\,\mathrm K$. The question does not specify temperature, so the result should also be understood through its temperature-dependent formula. The photocurrent density is
$$
J_{\rm ph}=q\Phi=(1.60218\times10^{-19})(2.88\times10^{21})
=461.43\,\mathrm{A\,m^{-2}}.
$$
At open circuit the diode current balances that photocurrent:
$$
\boxed{V_{\rm oc}=\frac{k_BT}{q}\ln\left(1+\frac{J_{\rm ph}}{J_0}\right)
=0.596\,\mathrm V\quad (298\,\mathrm K).}
$$
Using $300\,\mathrm K$ gives approximately $0.600\,\mathrm V$. These are current densities, so no device area is needed. Incomplete collection would replace $J_{\rm ph}$ by the collected current and reduce this estimate.
Without texturing, more incident photons generally escape by reflection, so the absorbed flux and $J_{\rm ph}$ decrease. Holding $J_0$ and temperature fixed, the voltage falls \b[logarithmically]: if absorption is multiplied by $f<1$ and $J_{\rm ph}\gg J_0$, then $\Delta V_{\rm oc}\simeq(k_BT/q)\ln f$. A real texture also changes surface area and recombination, so an unchanged $J_0$ is an assumption rather than a guarantee.
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