= Solution
Use the intended hot-ridge state and a linear law of <thermal expansion> about <mantle> <density> $\rho_m$ at $T_1$. Let $s$ be depth below the old seafloor. Its excess <density> is
$$
\delta\rho(s)=\rho_m\alpha\Delta T(1-s/a),\qquad 0<s<a,
$$
and its excess column <mass> is $\int_0^a\delta\rho\,ds=\rho_m\alpha\Delta T a/2$. <Isostasy> requires equal <pressure> at a compensation depth below both columns. Lowering the old seafloor by $h$ replaces rock by water, reducing column <mass> by $(\rho_m-\rho_w)h$. Therefore the <isostatic ridge elevation in a plate model> is
$$
\boxed{h=\frac{\rho_m\alpha\Delta T a}{2(\rho_m-\rho_w)}
=3.90\times10^3\,\mathrm m=3.9\,\mathrm{km}.}
$$
Here $\alpha\Delta T=0.052$; the result is a first-order estimate and neglects sediments, crustal-composition differences and flexural support. If one instead insists on the literal printed zero-age profile, its column is denser than the old plate and the formal relative elevation reverses sign, $-3.9\,\mathrm{km}$ to this order. \b[That profile cannot describe the elevated hot ridge posited by the question.]
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