For those that might be interested in the solution:
I have calculated the answer and I think this is the right answer:
Using the Hamiltonian from the question one can first calculate $\hat H|\psi>$:
For the kinetic part of the Hamiltonian: $$<\psi|\hat H_T|\psi>=-\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\frac{\hbar^2}{2m}\left( \frac{\partial^2}{\partial x_1^2}+\frac{\partial^2}{\partial x_2^2}\right)e^{-\frac{\left(x\space_1\space^2+x\space_2\space^2\right)}{a}}dx_1dx_2$$ $$<\psi|\hat H_T|\psi>=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\frac{\hbar^2}{2ma}\left( 2-(x_1^2+x_2^2)\right)e^{-\frac{\left(x\space_1\space^2+x\space_2\space^2\right)}{a}}dx_1dx_2$$ Solving this using Gaussian integrals gives: $$<\psi|\hat H_T|\psi>=\frac{\hbar^2\pi}{m}$$ For the potential part of the Hamiltonian without interaction term: $$<\psi|\hat H_V|\psi>=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\frac{1}{2}m\left( \omega_1^2x_1^2+\omega_2^2x_2^2\right)e^{-\frac{\left(x\space_1\space^2+x\space_2\space^2\right)}{a}}dx_1dx_2$$ Which gives: $$<\psi|\hat H_V|\psi>=\frac{m\pi a^2}{4}\left( \omega_1^2+\omega_2^2\right)$$ The last term is the interaction term: $$<\psi|\hat H_{V'}|\psi>=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}g\delta(x_1-x_2)e^{-\frac{\left(x\space_1\space^2+x\space_2\space^2\right)}{a}}dx_1dx_2$$ $$<\psi|\hat H_{V'}|\psi>=\int_{-\infty}^{+\infty}ge^{-2\frac{x\space_1\space^2}{a}}dx_1$$ (Note that I have used the delta function property for integrals).
Normalisation gives: $$<\psi|\psi>=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-\frac{\left(x\space_1\space^2+x\space_2\space^2\right)}{a}}dx_1dx_2$$ $$<\psi|\psi>=\pi a$$ The variational energy E(a) then becomes: $$E(a)=\frac{<\psi|\hat H|\psi>}{<\psi|\psi>}=\frac{\hbar^2}{ma}+g\frac{1}{\sqrt{2\pi a}}+\frac{ma}{4}\left( \omega_1^2+\omega_2^2\right)$$ The ground state energy can now be calculated by deriving E in respect to a: $\partial E/\partial a=0$ and filling in the given a back into the energy equation. 955