Let the thickness of the ice layer be $x$ at any given instant. The temperature at the ice-water interface is $0^\circ\text{C}$, and the outside air temperature is $-26^\circ\text{C}$. The rate of heat conduction $\frac{dQ}{dt}$ through the ice layer of cross-sectional area $A$ and thickness $x$ is given by Fourier's law of heat conduction: $\frac{dQ}{dt} = \frac{kA\Delta T}{x} = \frac{kA(0 - (-26))}{x} = \frac{26kA}{x}$ This heat is extracted from the water just below the ice, causing it to freeze. Let a small thickness $dx$ of ice form in a time interval $dt$. The mass of this newly formed ice is $dm = \text{volume} \times \text{density} = (A \cdot dx) \rho$. The heat released when this mass $dm$ of water freezes is $dQ = dm \cdot L = (A \rho dx)L$. The rate of heat release due to freezing is $\frac{dQ}{dt} = \rho A L \frac{dx}{dt}$. Equating the rate of heat release to the rate of heat conduction through the ice: $\rho A L \frac{dx}{dt} = \frac{26kA}{x}$ $\implies \frac{dx}{dt} = \frac{26k}{x\rho L}$ Therefore, the rate of increase of the thickness of the ice layer is $\frac{26k}{x\rho L}$.