Answer: \(\boxed{0.3mg}\).

Let the downward acceleration of the block be \(a_B\), and let the angular acceleration of the disc be \(\alpha\).

Because the string unwinds from the rim without slipping,

\[a_B=a\alpha.\]

For a uniform disc,

\[I=\frac12ma^2.\]

Taking moments about the centre of the disc gives

\[Ta=I\alpha.\]

So

\[Ta=\frac12ma^2\alpha.\]

Using \(a_B=a\alpha\), this becomes

\[T=\frac12ma_B.\]

Now resolve forces vertically on the block. Its weight is \(3mg\), and the vertical resistance is \(0.9mg\). Hence

\[3mg-0.9mg-T=3ma_B.\]

That is

\[2.1mg-T=3ma_B.\]

From \(T=\frac12ma_B\), we have \(a_B=\frac{2T}{m}\). Substitute this into the block equation:

\[2.1mg-T=3m\left(\frac{2T}{m}\right).\]

So

\[2.1mg-T=6T.\]

Therefore

\[7T=2.1mg,\]

and hence

\[\boxed{T=0.3mg}.\]