One answer is trivially "Yes" - fix some first-order sentence $\varphi$ which is provable from $X\Sigma_n$ but not $RCA_0$ alone and consider the formula $$\psi(x)\equiv (\varphi \vee \neg(x\in 0'))$$ (where "$x\in 0'$" is shorthand for the standard $\Sigma^0_1$ formula defining the Halting Problem). Then $RCA_0$ does not prove comprehension for $\psi$, but $X\Sigma_n$ does. Moreover, by choosing $\varphi$ to be a true $\Pi^0_1$ sentence, we can have $\psi$ be $\Pi^0_1$, so this gives examples as low in the arithmetic hierarchy as possible.

I suspect this is not what you mean. The problem is, it's not clear how to interpret your question. The $\omega$-models of $RCA_0$ and of $X\Sigma_n$ are of course the same, so there are no "true" sets of natural numbers whose existence requires induction.

Maybe you can clarify what you are asking?

EDIT: Another response which you might find more satisfactory:

Actually, induction axioms should be thought of as **anti**comprehension axioms - they show that definable cuts can't exist!

One instance of this is:

Let $M=(\omega_M, \mathbb{R}_M)$ be a model of $RCA_0$ such that, for some $\mathbb{R}'\supseteq\mathbb{R}_M$, we have $(\omega_M, \mathbb{R}')\models X\Sigma_n$. Then $M\models X\Sigma_n$.

Basically, the only way a model $M$ of $RCA_0$ could fail to be a model of $X\Sigma_n$ is if the first-order part of $M$ has a $\Sigma_n$-definable-from-parameters-in-$\mathbb{R}_M$ cut; but if $(\omega_M, \mathbb{R}')\models X\Sigma_n$ for some larger $\mathbb{R}'$, then this can't happen.

In particular, for any model $\omega_M$ of the first-order part $I\Sigma_n$ of $X\Sigma_n$, the structure $(\omega_M, D)$ - where $D$ is the class of $\Delta^0_1$ subsets of $\omega_M$ - is a model of $X\Sigma_n$. But $(\omega_M, D)$ is also the smallest model of $RCA_0$ with first-order part $\omega_M$!

So adding induction doesn't add sets.