= Solution
Take rightward translation as positive and counterclockwise <angular acceleration> $\alpha$ as positive. Let $f$ denote signed ground <friction>, positive to the right. The source's lower-tangent winding makes the <tension> exert counterclockwise <torque> $Tr$. For a horizontal pull, translation and rotation obey
$$
Ma=T+f,\qquad I\alpha=Tr+Rf,\qquad a=-R\alpha.
$$
The final relation is the <rolling without slipping> constraint. Eliminating $f$ and $\alpha$ gives
$$
\boxed{a=\frac{TR(R-r)}{I+MR^2}>0,\qquad
f=-T\frac{I+MrR}{I+MR^2}.}
$$
Thus the requested magnitude of <static friction> is
$$
\boxed{F=|f|=T\frac{I+MrR}{I+MR^2}.}
$$
It acts leftward. Released from rest, the spool accelerates \b[rightward, toward the pulling hand], and rotates clockwise. There is no contradiction between leftward <friction> and rightward motion: <static friction> prevents slipping of the contact point, and the horizontal <tension> is larger than $|f|$ because $r<R$.
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