One uparrow in Knuth's notation is just regular exponentiation, so you are asking about $N=(10^{10^{100}})^{10^{10^{100}}}=10^{(10^{100}\cdot {10^{10^{100}}})}$ which has $10^{100}\cdot {10^{10^{100}}}+1$ digits.
This is actually not such a large number. If we take $\log_{10} (\log_{10} (\log_{10} N)))$ we get about $100$. In this answer I consider $3 \uparrow \uparrow \uparrow 3$ and find it takes $7625597484985$ applications of the $\log$ to make the number handy. We define $\log^*$ as the number of applications of $\log$ to make a number handy, so we have $\log^* N=4$, but $\log^*3 \uparrow \uparrow \uparrow 3=7625597484985$ so it takes $\log \log^*3 \uparrow \uparrow \uparrow 3$ to get handy. Bigger numbers will need $\log^{**}$, the number of times you need to apply $\log^*$ and so on.