A simple upper bound:

>! $7^2 = 49$ is just under $50$, so $7^{100} = (7^2)^{50}$ is somewhat less than $50^{50} = (100/2)^{50}$.

>! $2^{10} = 1024$ is just over $1000 = 10^3$, so $2^{50}$ is somewhat over $10^{15}$.

>! So $7^{50} < (100^{50}) / (10^{15}) = ((10^2)^{50}) / (10^{15}) = (10^{100}) / (10^{15}) = 10^{85} = 1\mathrm{e}85$.

And a semi-simple lower bound:

>! $7^6 = 117649$ is somewhat more than $10^5$, so $7^{96} = (7^6)^{16}$ is somewhat more than $(10^5)^{16} = 10^{80}$.

>! $7^{100} = (7^{96}) * (7^4)$ is somewhat more than $10^{80} * 2401 = 2.40 1\mathrm{e}83$.

Or, borrowing an answer from the previous problem: If you have

>! a log table giving $log_{10}(7) \approx 0.845$

then you can determine that

>! $log_{10}(7^{100}) = 100 * log_{10}(7) \approx 84.5$

which produces an estimate of

>! $\sqrt{10} * 10^{84} ~= 3.162\mathrm{e}84$