A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will

According to Gauss law the total flux through a surface is given by $\mathrm{\phi }=\frac{{\mathrm{q}}_{\mathrm{en}}}{{\mathrm{\epsilon }}_{\mathrm{o}}}$.

This is independent of the radius of the surface, so the outward electric flux will remain the same.