Two governing principles are:

(i) Electric field inside all the metallic plates should be zero

(ii) Net charge on the plates (excess + induced) should be equal to the excess charge as stated in the problem statement

So, electric field inside plate $1$ should be zero implies

$\sigma_a = \sigma_b + \sigma_c + \sigma_d + \sigma_e + \sigma_f +\sigma_g + \sigma_h + \sigma_i + \sigma_j$ $\cdots$ (1)

electric field inside plate $2$ should be zero gives us

$\sigma_a + \sigma_b + \sigma_c = \sigma_d + \sigma_e + \sigma_f +\sigma_g + \sigma_h + \sigma_i + \sigma_j$ $\cdots$ (2)

electric field inside plate $3$ should be zero gives us

$\sigma_a + \sigma_b + \sigma_c + \sigma_d + \sigma_e = \sigma_f +\sigma_g + \sigma_h + \sigma_i + \sigma_j$ $\cdots$ (3)

electric field inside plate $4$ should be zero gives us

$\sigma_a + \sigma_b + \sigma_c + \sigma_d + \sigma_e + \sigma_f +\sigma_g = \sigma_h + \sigma_i + \sigma_j$ $\cdots$ (4)

electric field inside plate $5$ should be zero gives us

$\sigma_a + \sigma_b + \sigma_c + \sigma_d + \sigma_e + \sigma_f +\sigma_g + \sigma_h + \sigma_i = \sigma_j$ $\cdots$ (5)

And conservation of charge would give us

$\sigma_a + \sigma_b = \cfrac{Q}{A}$ $\cdots$ (6)

$\sigma_c + \sigma_d = \cfrac{2Q}{A}$ $\cdots$ (7)

$\sigma_e + \sigma_f = \cfrac{3Q}{A}$ $\cdots$ (8)

$\sigma_g + \sigma_h = \cfrac{4Q}{A}$ $\cdots$ (9)

$\sigma_i + \sigma_j = \cfrac{5Q}{A}$ $\cdots$ (10)

Adding (1) and (5), we get $\sigma_a = \sigma_j$

Subtracting (2) from (1) we get $\sigma_b = – \sigma_c$

Subtracting (3) from (2) we get $\sigma_d = – \sigma_e$

Subtracting (4) from (3) we get $\sigma_f = – \sigma_g$

Subtracting (5) from (4) we get $\sigma_h = – \sigma_i$