Answer: 896

Solution:

Exponentiating the first equation twice: $\log_a(\log_a 2) + \log_a 24 - 128$ $= a^{128}$ becomes $\log_a(24 \log_a 2) = 128 + a^{128},$ so $24 \log_a 2 = a^{128} \cdot a^{a^{128}},$ i.e. $2^{24} = a^{\left(a^{128} \cdot a^{a^{128}}\right)}.$ Setting $t = a^{a^{128}},$ the right side is $t^t$ and the left side is $\left(2^3\right)^{2^3},$ so by the strict monotonicity of $t^t$ we get $a^{a^{128}} = 8.$ Writing $c = \log_2 a \gt 0,$ this says $c \cdot 2^{128c} = 3,$ which is increasing in $c$ and satisfied by $c = \frac{3}{64}:$ indeed $\frac{3}{64} \cdot 2^6 = 3.$ So $a = 2^{3/64}.$

The second equation gives $x = a^{a^{256}}.$ Here $a^{256} = 2^{256 \cdot 3/64} = 2^{12} = 4096,$ so $$x = a^{4096} = 2^{4096 \cdot 3/64} = 2^{192}.$$

Clearly $2^{192} \equiv 0 \pmod{8}.$ By Euler's theorem $2^{100} \equiv 1 \pmod{125},$ so $2^{192} \equiv 2^{-8} \pmod{125},$ the inverse of $256 \equiv 6.$ Since $6 \cdot 21 = 126 \equiv 1 \pmod{125},$ we get $2^{192} \equiv 21 \pmod{125}.$ The unique residue mod $1000$ that is $0$ mod $8$ and $21$ mod $125$ is $896.$