Dimensional Analysis:

$ \mathrm{T} \propto \mathrm{m}^{\mathrm{x}} \mathrm{G}^{\mathrm{y}} \mathrm{a}^{\mathrm{z}} $

Using dimensional analysis for gravitational interactions, we have:

$ \mathrm{T} \propto \mathrm{M}^{\mathrm{x}} G^{\mathrm{y}} a^{\mathrm{z}} $

Where $ G $ has dimensions $\left[\mathrm{M}^{-1} \mathrm{L}^3 \mathrm{T}^{-2}\right]$.

Solving for Exponents:

$ \mathrm{T} \propto \mathrm{M}^{\mathrm{x-y}} \mathrm{L}^{3y+z} \mathrm{T}^{-2y} $

Equating dimensions, we solve:

$ \mathrm{x}-\mathrm{y}=0 \Rightarrow \mathrm{x}=\mathrm{y} $

$ -2\mathrm{y}=1 \Rightarrow \mathrm{y}=-\frac{1}{2}, \mathrm{x}=-\frac{1}{2} $

$ 3\mathrm{y}+\mathrm{z}=0 \implies \mathrm{z}=-3\mathrm{y}=\frac{3}{2} $

Time Proportionality:

$ \mathrm{T} \propto \mathrm{~m}^{-1/2} \mathrm{G}^{-1/2} \mathrm{a}^{3/2} $

Which simplifies to:

$ \mathrm{T} \propto \left(\frac{a^3}{m}\right)^{1/2} $

Applying New Conditions:

When the side length becomes $2a$ and mass becomes $2m$:

$ \mathrm{T} = 4 \times \left(\frac{(2a)^3}{2m}\right)^{1/2} $

Simplifying:

$ \mathrm{T} = 4 \times \left(\frac{8a^3}{2m}\right)^{1/2} = 4 \times (4)^{1/2} = 8 \text{ seconds} $

Thus, the spheres will collide after 8 seconds under the new conditions.