= Solution
The extra <ocean heat flux> supplies energy per unit area
$$
\Delta Q=6\,\mathrm{W\,m^{-2}}(275\times86400\,\mathrm s)=1.4256\times10^8\,\mathrm{J\,m^{-2}}.
$$
In a fixed heat-removal estimate, this offsets the <latent heat of fusion> that would otherwise form thickness $\Delta H=\Delta Q/(\rho_iL_f)$. Using a standard ice <mass density> $\rho_i\simeq917\,\mathrm{kg\,m^{-3}}$, an additional material assumption,
$$
\boxed{\Delta H\simeq0.46\,\mathrm m,\qquad H_1\simeq1.5-0.46\simeq1.04\,\mathrm m.}
$$
For any other assumed ice density, retain $\Delta H=1.4256\times10^8/(336000\rho_i)$ metres. This is the latent-heat equivalent of the added energy, not an exact coupled growth calculation. Thinner ice conducts heat more readily, so the atmospheric heat-removal budget can respond; a full <sea ice thermodynamics> model, such as the <single-category sea-ice thermodynamic model>, includes that compensation.
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