$f_0(x) = x^2$
and $f_n(x) = f_{n1}(f_{n1}(x))$
for $n > 0.$
How many digits are in the number $f_n(10)$
for a given $n \ge 0?$

Have you tried finding the general formula for
$f_n(x)?$

How many digits does
$10^n$
have?

We can work out the general formula for
$f_n(x)$
and then substitute$x=10$
:$$\begin{aligned} f_1(x)&=(x^2)^2=x^4 \\ f_2(x)&=(x^4)^4=x^{16} \\ f_3(x)&=(x^{16})^{16}=x^{256} \\ &\ldots \\ f_n(x)&=x^{2^{2^n}} \\ \end{aligned}$$
We could show that our “guess” satisfies the recurrence relation:
$$\begin{aligned} f_n(x)&=f_{n1}(f_{n1}(x)) \\ &=\big(x^{2^{2^{n1}}}\big)^{2^{2^{n1}}} \\ &= x^{2^{2^{n1}} \cdot {2^{2^{n1}}}} \\ &=x^{2^{2^n}} \\ \end{aligned}$$
The answer is thus
$2^{2^n}+1$
digits.