Let $\frac{202^{7^m}+1}{v_7\left(202^{7^m}+1\right)}=a$. Note that $$v_7(202^{7^m}+1)=v_7(202+1)+v_7(7^m)=1+m.$$Thus, $7^i\cdot a$ with $i=1,2,\dots,m$ are all positive divisors of $202^{7^m}+1$ which has exactly the same set of prime divisors with $202^{7^m}+1$. And so $\rho\left(202^{7^m}+1\right)\geq m$ for every positive integers $m$.

If $n=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_k^{\alpha_k}$, then we have that $\rho (n)= \alpha_1 \alpha_2 \dots \alpha_k$. Notice that by LTE if $n$ is odd we have that $v_7 (202^n+1)=1+v_7(n)$, thus set $n=7^k$. Also notice that $\rho (202^{7^k}+1) \geq 1+k \geq m$, thus set $k >m$, and we are done

Or $U_29$ or any prime which devise one of the numbers$202^k+1$

$n=7^{m-1}$ $$VERY EASY PROBLEM $$