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Emil's answer gives you the ordinals you are asking for, but it may be worth to add a complementary remark: With restricted induction, one needs to be slightly careful about how the ordinals are computed.

For example, the provably recursive functions of $I\Sigma^0_1$ are precisely the primitive recursive functions.

However, these are precisely the $\omega^2$-recursive functions, i.e., those that can be proved total using the infinitary proof-system $\vdash^{\omega^2}_0$ of Tait "Normal derivability in classical logic", in "The syntax and semantics of infinitary languages", Lecture Notes in Mathematics 72, Springer, pp. 204-236.

Given $f:{\mathbb N}\to{\mathbb N}$, let $E(f)$ consist of all those functions "explicitly definable" using 0,1,$f$,$+$, restricted subtraction, and bounded sums and products.

There are several fast-growing hierarchies of recursive functions one uses to analyse fragments of arithmetic. The functions $B_\alpha$ are defined inductively: $B_0(n)=n+1$, $B_{\alpha+1}(n)=B_\alpha(B_\alpha(n))$, $B_\lambda(n)=B_{\lambda_n}(n)$ for $\lambda$ limit, where the $\lambda_n$ are the "natural" strictly increasing sequence of ordinals converging to $\lambda$.

The functions $F_\alpha$ are defined similarly, except that $F_{\alpha+1}(n)=F^{n+1}_\alpha(n)$, where the superindex denotes iterated composition ($n+1$ times). This is the sequence of functions most used in this context.

The functions of the Hardy hierarchy are defined by $H_0(n)=n$, $H_{\alpha+1}(n)=H_\alpha(n+1)$, and $H_\lambda(n)=H_{\lambda_n}(n)$.

Then the primitive recursive functions are precisely the functions in $$\bigcup_{\alpha\prec\omega^2}E(B_\alpha)=\bigcup_{\alpha\prec\omega}E(F_\alpha)=\bigcup_{\alpha\prec\omega^\omega}E(H_\alpha);$$ here, $\prec$ is a partial subordering of the ordinals, but at this level we may identify it with the usual $\le$.\lt$. All this is discussed in great detail in the nice paper by Fairtlough and Wainer, "Hierarchies of provably recursive functions", in "Handbook of Proof Theory", Elsevier, pp. 149-207. 1 Emil's answer gives you the ordinals you are asking for, but it may be worth to add a complementary remark: With restricted induction, one needs to be slightly careful about how the ordinals are computed. For example, the provably recursive functions of$I\Sigma^0_1$are precisely the primitive recursive functions. However, these are precisely the$\omega^2$-recursive functions, i.e., those that can be proved total using the infinitary proof-system$\vdash^{\omega^2}_0$of Tait "Normal derivability in classical logic", in "The syntax and semantics of infinitary languages", Lecture Notes in Mathematics 72, Springer, pp. 204-236. Given$f:{\mathbb N}\to{\mathbb N}$, let$E(f)$consist of all those functions "explicitly definable" using 0,1,$f$,$+$, restricted subtraction, and bounded sums and products. There are several fast-growing hierarchies of recursive functions one uses to analyse fragments of arithmetic. The functions$B_\alpha$are defined inductively:$B_0(n)=n+1$,$B_{\alpha+1}(n)=B_\alpha(B_\alpha(n))$,$B_\lambda(n)=B_{\lambda_n}(n)$for$\lambda$limit, where the$\lambda_n$are the "natural" strictly increasing sequence of ordinals converging to$\lambda$. The functions$F_\alpha$are defined similarly, except that$F_{\alpha+1}(n)=F^{n+1}_\alpha(n)$, where the superindex denotes iterated composition ($n+1$times). This is the sequence of functions most used in this context. The functions of the Hardy hierarchy are defined by$H_0(n)=n$,$H_{\alpha+1}(n)=H_\alpha(n+1)$, and$H_\lambda(n)=H_{\lambda_n}(n)$. Then the primitive recursive functions are precisely the functions in $$\bigcup_{\alpha\prec\omega^2}E(B_\alpha)=\bigcup_{\alpha\prec\omega}E(F_\alpha)=\bigcup_{\alpha\prec\omega^\omega}E(H_\alpha);$$ here,$\prec$is a partial subordering of the ordinals, but at this level we may identify it with the usual$\le\$.

All this is discussed in great detail in the nice paper by Fairtlough and Wainer, "Hierarchies of provably recursive functions", in "Handbook of Proof Theory", Elsevier, pp. 149-207.