= Solution
Electrical contact equalizes <electric potential> and preserves total <electric charge>. \b[The usual isolated-sphere approximation] assigns the two <conducting spheres> the <capacitances> $4\pi\epsilon_0R$ and $8\pi\epsilon_0R$. It then gives
$$
\frac{Q_R}{R}=\frac{Q_{2R}}{2R},\qquad Q_R+Q_{2R}=Q,
\qquad \boxed{Q_R\simeq Q/3,\quad Q_{2R}\simeq2Q/3.}
$$
These are accurate when the <conducting spheres> are far apart and joined by a thin wire. \b[Literal touching changes the answer:] each <conducting sphere> polarizes the other, so its <electric potential> is not its isolated value. The source does not specify an approximation which would remove this effect.
For completeness, the exact result for ideal externally tangent <conducting spheres> can be obtained by the <method of images>. Let their radii be $a,b$, their centre separation $D=a+b$, and their common <electric potential> $V$. Begin with central <image charges> $4\pi\epsilon_0Va$ and $4\pi\epsilon_0Vb$. Reflect each external <image charge> $q'$ at centre-distance $s$ into the other <conducting sphere> of radius $a$: the new <image charge> is $-a q'/s$ at centre-distance $a^2/s$ toward the original. Repeat, alternately, in the two <conducting spheres>. Pairing successive opposite-sign terms gives the convergent net <electric charges>
$$
\begin{aligned}
Q_a&=4\pi\epsilon_0Vab\sum_{n=0}^{\infty}\left(\frac1{b+nD}-\frac1{(n+1)D}\right),\\
Q_b&=4\pi\epsilon_0Vab\sum_{n=0}^{\infty}\left(\frac1{a+nD}-\frac1{(n+1)D}\right).
\end{aligned}
$$
Each reflection cancels a nonconstant surface <electric potential>; the remaining central term fixes $V$. The total internal <image charge> equals the exterior <electric flux> divided by $1/\epsilon_0$, so these sums give the physical net <electric charges>.
For $a=R,b=2R$, multiply both dimensionless sums by the common factor $4\pi\epsilon_0V(2R/3)$. Their values follow by writing reciprocals as <integrals> and setting $t=u^3$:
$$
\begin{aligned}
S_a&=\sum_{n=0}^{\infty}\left(\frac1{n+2/3}-\frac1{n+1}\right)=3\int_0^1\frac{u\,du}{1+u+u^2}=\frac32\ln3-\frac{\pi}{2\sqrt3},\\
S_b&=\sum_{n=0}^{\infty}\left(\frac1{n+1/3}-\frac1{n+1}\right)=3\int_0^1\frac{(1+u)\,du}{1+u+u^2}=\frac32\ln3+\frac{\pi}{2\sqrt3}.
\end{aligned}
$$
Dividing by $S_a+S_b$ enforces <conservation of electric charge> and yields
$$
\boxed{Q_R=Q\left(\frac12-\frac{\pi}{6\sqrt3\ln3}\right)\simeq0.22483Q,\qquad Q_{2R}=Q\left(\frac12+\frac{\pi}{6\sqrt3\ln3}\right)\simeq0.77517Q.}
$$
This concrete touching-sphere result disproves applying the isolated $1:2$ sharing rule to literal contact.
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