For solving the simplified one-dimensional Fick's equation
$$ \frac{\partial c}{\partial t} = D \frac{\partial^2 c}{\partial x^2} $$
with the common boundary conditions, the Laplace transform results in
$$c = \frac{A}{t^{1/2}}exp(-x^2/4Dt)$$
but if having a two-dimensional diffusion (assume diffusion over a flat surface) as $$ \frac{\partial c}{\partial t} = D \Bigl(\frac{\partial^2 c}{\partial x^2} + \frac{\partial^2 c}{\partial y^2}\Bigl)$$
then, how will be the Laplace transform to solve the two-dimensional form of the Fick's second law?
You can look up solutions to the heat equation in higher dimensions.
The equation $$ \partial_t u(\vec{x},t) = \kappa\sum_i \partial_{ii} u(\vec{x},t) $$ has Green's function (fundamental solution) $$ \Psi(\vec{x},t) = \frac{1}{\sqrt{(4\pi \kappa t)^n}}\exp\left(\frac{-[\vec{x}\cdot\vec{x}]}{4\kappa t}\right) $$ So that given initial conditions $u(\vec{x},0)=\mathcal{I}(\vec{x})$, the solution is $$ u(\vec{x},t)=\int_{\mathbb{R}^n} \Psi(\vec{x}-\vec{y},t)\;\mathcal{I}(\vec{y})\, d\vec{y} $$ Using the notation of the heat kernel instead, we can write: $$ \mathfrak{K}_t(\vec{x},\vec{y})=\frac{1}{\sqrt{(4\pi \kappa t)^n}}\exp\left(\frac{-||\vec{x}-\vec{y} ||^2_2}{4\kappa t}\right) $$ so it has solution: $$ u(\vec{x},t)=\int_{\mathbb{R}^n} \mathfrak{K}_t(\vec{x},\vec{y})\;\mathcal{I}(\vec{y})\, d\vec{y} $$ So if $n=2$ and $\mathcal{I}(\vec{y})=\delta(\vec{y})$, we get: $$ u(x,y,t) = \frac{1}{{4\pi \kappa t}}\exp\left(\frac{-[x^2+y^2]}{4\kappa t}\right) $$