Two metallic spheres which have equal charges but their radii are different are made to touch

As Martin Nicholson's answer describes, doing this properly is complicated. However if you are prepared to accept a rather arm waving and approximate argument we can easily see intuitively why the larger sphere carries more charge.

We can calculate the potential at the surface of a charged sphere like this. A spherically symmetric charged body behaves as a point charge. By this I mean that the electric potential outside the sphere is the same as for a point charge at the centre of the sphere. This is a consequence of Gauss's law.

So suppose we have a sphere of radius $R$ and charge $Q$, the potential at a distance $r > R$ from the centre of the sphere is just:

$$ V = -\frac{1}{4\pi\epsilon_0}\frac{Q}{r} $$

And the voltage at the surface of there sphere is therefore:

$$ V_{surface} = -\frac{1}{4\pi\epsilon_0}\frac{Q}{R} \tag{1} $$

Now consider your two spheres of radii $R_1$ and $R_2$, and charges $Q_1$ and $Q_2$. Suppose the spheres are far enough apart that their fields don't affect each other e.g. we connect them with a long wire. When the spheres are connected together they will have the same potential because current can flow between them until the potential difference is zero. That means:

$$ V_1 = V_2 $$

and using equation (1) for the surface voltage we find:

$$ -\frac{1}{4\pi\epsilon_0}\frac{Q_1}{R_1} = -\frac{1}{4\pi\epsilon_0}\frac{Q_2}{R_2} $$

or more simply:

$$ \frac{Q_1}{R_1} = \frac{Q_2}{R_2} $$

And we can rearrange this to get $Q_1$ we get:

$$ Q_1 = Q_2\frac{R_1}{R_2} $$

and since we know $R_1 > R_2$ it follows that $Q_1 > Q_2$.

Now this only applies when the spheres are far apart, but it will approximately apply at closer distances, so we expect the larger sphere will carry more charge even when the spheres are close together.

The ratio of $Q/V$ is called the capacitance, or in this case it would be more precise to call it the self capacitance, and the capacitance of a conducting sphere is:

$$ C = \frac{Q}{V} = 4\pi\epsilon_0 R $$

Another way of answering your question is to point out that the capacitance of a large sphere is greater than the capacitance of a small sphere, so when the voltages are equal the large sphere will contain a greater charge.

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