Use Fe415 and M20 grade concrete

$$\dfrac {ly}{lx}=\dfrac {5.75+0.23}{4+0.23}=\dfrac {5.98}{4.23}=1.41 < 2$$

Therefore it is a two way slab

$d=\dfrac {\text {short effective span}}{\text {S⁄D ratio ×M.F}}= \dfrac {4230}{20×1.4}=151.07mm ≈160\\ \text {D=d+effective cover } \\ =160+25=185mm\\ \text { Load Calculation } \\ D.L = D×25=0.185×25=4.625kN/m^2 \\ L.L =3.5kN/m^2 \\ F.F = 1 kN/m^2 \\ Total =9.125 kN/m^2 \\ \text {Factored load }(w_d) =9.125 ×1..5=13.68 kN/m\\ \dfrac {ly}{lx}=1.41 $

$$M_{ux}=α_x wdlx^2 $$

$=0.0998×13.68×4.23^2 \\ =24.42 kNm \\ M_{uy}= α_y wdlx^2 \\ =0.138=0.0509×13.68×4.23^2 \\ =12.45 kNm \\ M_{u\space max}=0.138f_{ck} bd^2\\ =0.138×20×1000×160^2 \\ =70.65 kNm \gt M_{ux}\space \& \space M_{uy} \\ Astx= \dfrac {0.5×20×1000×160}{415}×[1-\sqrt{1-\dfrac {4.6×24.42×10^6}{20×1000×160^2 }}] \\ =449.09 mm^2 \\ \text {Assume 8mm∅ bars} ∅_x=∅_y=8d_y=160-\dfrac 82-\dfrac 82=152mm\\ Asty= \dfrac {0.5×20×1000×152}{415}×[1-\sqrt{1-\dfrac {4.6×12.45×10^6}{20×1000×152^2 }}]\\ =234.47 mm^2 \\ Astmin=\dfrac {0.12}{100} b×D=\dfrac {0.12}{100}×1000×185=222 mm^2 \lt Astx\space \&\space Asty \\ \text { * Main steel in x-direction (8mm∅) } spacing =\dfrac {b×c/s\space area}{Asty}=\dfrac {1000×\pi/4 ×8^2}{449.09}=111.92 ≈100mm\\ \text {* Main steel in y-direction (8mm∅) } spacing =\dfrac {b×c/s\space area}{Asty}=\dfrac {1000×\pi/4 ×8^2}{234.47}=214.37 ≈200mm$