$\left(x^{2}+\frac{a}{x^{3}}\right)^{10}$

As we know that the general term $\left(T_{r+1}\right)$ of the expression $(a+b)^{n}$ is-

$T_{r+1}={ }_{n} C_{r} a^{n-r} b^{n}$

Therefore, in the given expansion $\left(x^{2}+\frac{a}{x^{3}}\right)^{10}$ $a=x^{2}$

$b=\frac{a}{x^{3}}$

$n=10$

$\therefore$ general term of the given expression-

$T_{r+1}={ }_{10} C_{r} x^{20-2 r} \frac{a^{r}}{x^{3 r}}$

$\Rightarrow T_{r+1}={ }_{10} C_{r} x^{20-5 r} a^{r}$

For coefficient of $x^{5}-$

$20-5 r=5$

$5 r=15 \Rightarrow r=3$

$\therefore T_{4}={ }_{10} C_{3} x^{5} a^{3}$

For coefficient of $x^{15}-$

$20-5 r=15$

$5 r=5 \Rightarrow r=1$

$\therefore T_{2}={ }_{10} C_{1} x^{15} a$

Given that the coefficient of $x^{5}$ is equal to the coefficient of $x^{15}$, i.e.,

${ }_{10} C_{3} a^{3}={ }_{10} C_{1} a \ldots$

As we know that,

${ }_{n} C_{r}=\frac{n !}{r !(n-r) !}$

$\therefore{ }_{10} C_{3}=\frac{10 !}{3 !(10-3) !}=120$

${ }_{10} C_{1}=\frac{10 !}{1 !(10-1) !}=10$

Now from $e q^{n}(1)$, we have

$120 \times a^{3}=10 a$

$\Rightarrow a^{2}=\frac{1}{12}$

$\Rightarrow a=\pm \frac{1}{2 \sqrt{3}}$

As we have to find positive value of $a$. $\therefore a=\frac{1}{2 \sqrt{3}}$