Answer:
The value is
![q_o = Q](https://img.qammunity.org/2021/formulas/physics/college/60oanongrcy05xdzgf4yo5b8vm20cwddkd.png)
Step-by-step explanation:
From the question we are told that
The net charge is
![Q = + 3Q](https://img.qammunity.org/2021/formulas/physics/college/y35j6t3n5kne7jphsksr8bebm21g4d1m0d.png)
The charge place on the inside of the cavity is
![q = -2Q](https://img.qammunity.org/2021/formulas/physics/college/eqsquff9u5l3n1ztlc3wmlpu5dus46420e.png)
Since we are told from the question that small charge placed inside the cavity is isolated from the conductor
Then it implies that the electric flux is Zero
Which mean that the charge place within the conductor + the charge on the inner region of the conductor = 0
i.e
![q + q_i = 0](https://img.qammunity.org/2021/formulas/physics/college/xm79j5m3gx7uzgklh0nhk94y0mi4z7deu3.png)
=>
![-2Q + q_i = 0](https://img.qammunity.org/2021/formulas/physics/college/s6qveiw6ey3j61uqdmykrxhg0lkddcmr6s.png)
=>
![q_i = 2Q](https://img.qammunity.org/2021/formulas/physics/college/u5yzchm7qlovewa3b5g5slgtkkxn5ld9ab.png)
Now the net charge on the conductor is mathematically represented as
![Q = q_o + q_i](https://img.qammunity.org/2021/formulas/physics/college/l9m7cgfii87eogj4wpywixr3ji3ib64u3b.png)
Here
is the charge on the outer surface
So
![3Q = q_o + 2Q](https://img.qammunity.org/2021/formulas/physics/college/b2eh8ctvc5efe50vlqde1cceaclvoj8ln9.png)
=>
![q_o = Q](https://img.qammunity.org/2021/formulas/physics/college/60oanongrcy05xdzgf4yo5b8vm20cwddkd.png)