Total number of $5$ string sets is $\binom{26^{10}}{5}$
Out of which, total $\binom{26^{10}-1}{4}$ sets will contain password.
So required probability that his guessed set contains the password will be $\frac{\binom{26^{10}-1}{4}}{\binom{26^{10}}{5}}$.
And if you simplify $\frac{\binom{n-1}{r-1}}{\binom{n}{r}}$, it will be $\frac{r}{n}$ $\rightarrow$ $\frac{5}{26^{10}}$.
$\rightarrow$ So, Option $A$