Answer:
Approximately [tex]10.7\; {\rm N}[/tex] (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)
Explanation:
Let the magnitude of this tension be [tex]F_{\text{tension}}[/tex].
In block [tex]\text{A}[/tex], this tension points towards the edge of the table and is unbalanced. Hence, the net force on block [tex]\text{A}\![/tex] would be equal to the tension, [tex]F_{\text{tension}}[/tex]. The acceleration of block [tex]\text{A}\!\![/tex] would be [tex](F_\text{tension}) / (m_{\text{A}})[/tex].
In block [tex]\text{B}[/tex], this tension points upwards and acts against the weight [tex]m_{\text{B}}\, g[/tex] of this block. Hence, the net force on block [tex]\text{B}\![/tex] would be [tex](m\, g - F_{\text{tension}})[/tex]. The acceleration of block [tex]\text{B}\!\![/tex] would be [tex](m\, g - F_{\text{tension}}) / (m_{\text{B}})[/tex].
If the light string between the two blocks is inelastic, the acceleration of the two blocks should be equal in magnitude. Hence:
[tex]\displaystyle \frac{F_\text{tension}}{m_{\text{A}}} = \frac{m_{\text{B}}\, g - F_{\text{tension}}}{m_{\text{B}}}[/tex].
Rearrange this equation and solve for [tex]F_{\text{tension}}[/tex]:
[tex]\displaystyle \frac{F_\text{tension}}{m_{\text{A}}} = g - \frac{F_{\text{tension}}}{m_{\text{B}}}[/tex].
[tex]\begin{aligned} F_{\text{tension}} &= \frac{g}{(1/m_{\text{A}}) + (1 / m_{\text{B}})} \\ &= \frac{9.81\; {\rm m\cdot s^{-2}}}{(1 / (9\; {\rm kg})) + (1 / (1.24\; {\rm kg}))} \\ &\approx 10.7\; {\rm N}\end{aligned}[/tex].
