Use $M20/Fe415.$ Draw neat sketch showing reinforcement details.

$$l_y=10+0.23=10.23 $$ $ l_x=2.5+0.23=2.73 \\ Here\space \dfrac {l_y}{l_x}= \dfrac {10.23}{2.73}=3.74\gt2 \\ \text {Hence one way slab. }\\ d= \dfrac{\text { short effective span}}{\text{S⁄D ratio ×M.F}}=\dfrac {2.73 ×1000}{20×1.4} \\ d= 97.5mm ≈ 100mm \\ D= d+25 =125mm \\ \text {Load Calculation}\\ D.L = D×25=0.125×25=3.125 kN/m^2 \\ F.F = 1kN/m^2 \\ L.L = 3kN/m^2 \\ W=7.125 kN/m^2 \\ \text {Factored load }(w_d) = 1.5×7.125=10.68kN/m^2 \\ Now \space M_u=\dfrac {wdlx^2}8=\dfrac {10.68×2.73^2}8=9.94 kNm \\ M_{u\space max}=0.138 f_{ck} bd^2 \\ = 0.138×20×1000×100^2=27.6 kNm \\ M_{u\space max} \gt M_u \space \space Safe \\ A_{st}=(\dfrac{0.5f_{ck} bd}{f_y} )×[1-\sqrt{1-\dfrac{4.6M_o}{f_{ck} bd^2 }}] \\ = (\dfrac {0.5×20×1000×100}{415})×[1-\sqrt{1-\dfrac {4.6×9.94×10^6}{20×1000×100^2 }}] \\ A_{st}=293.29 mm^2 \\ A_{st\space min}=\dfrac {0.12bd}{100}=\dfrac {0.12}{100 }×1000×125 \\ = 150mm \lt A_{st}\space \space Safe \\ \text { * Main steel of 10mm∅ }\\ Spacing :- S_1= \dfrac {b×c⁄s\space area}{A_{st }} \\ = \dfrac {1000×\pi⁄4×10^2}{293.92 }=267.07≈250mm \\ S_2=3d=3×100=300mm \\ S_3=300mm \\ \text { Provide 10mm∅ @ 250mm c/c } \\ \text {* Distribution steel of 8mm∅ } \\ Spacing :S_1=\dfrac {b×c⁄s \space area}{A_{st} } \\ =\dfrac {1000×\pi⁄4 ×8^2}{150}=333.33mm ≈325mm\\ S_2=5d=5×100=500mm \\ S_3=450mm \\ \text {Provide 8mm∅ @ 325mm c/c }$

$$\text {* Check for shear: (At support) }$$ $A_{st\space support}=\dfrac {100×\pi⁄4 ×10^2}{500}=157mm^2 \\ pt=\dfrac {100×A_{st\space support}}{b×d}= \dfrac {100×157}{1000×100}=0.157\% \\ pt=0.157 \% \hspace {1cm} Z_{uc}=0.285N/mm^2\space \& \space k=1.3 \\ V_{uc}=k×Z_{uc}×b×d=(1.3×0.285×1000×100)\\ V_{uc}=37.05kN \\ V_{uD}=\dfrac {wd×l_x}2= \dfrac {10.69×2.73}2=14.59kN \\V_{uc} \gt V_{uD} \space\space safe \\ \text {* Check for deflection:- (At midspan) } \\ Ast=\dfrac {1000×\pi⁄4 ×10^2}{250}=314mm^2 \\ pt=\dfrac {100×Ast}{b×d}=\dfrac {100×314}{1000×100}=0.314\% \\ f_s=0.58×f_y× \dfrac {Astr}{Astp}=0.58×415×\dfrac {293.92}{314}\\ f_s=225.3N/mm^2 \\ M.F =2 [\text { from graph pg }-38 IS:456] \\ D=\dfrac {2730}{20×2×1.6}=42.65mm\lt100mm \text { Therefore Safe} \\ \text {* Check for development length: } \\ L_d=\dfrac {0.87f_y∅}{4Zbd} = \dfrac {0.87×414×10}{4×(1.2×1.6)}=470.11mm \\ M_o= \dfrac {M_o}2=\dfrac {9.96}2=4.98kNm \\ V=V_{uD}=14.59kN \\ L_o=bs⁄2-25+3d=\dfrac {250}2-25+(3×10)=120mm \\ \dfrac{1.3M_o}V+L_o=(\dfrac {1.3×4.98×10^6}{14.59×10^3 }+120)=563.72mm\\ 470.11\lt563.72mm \text { Therefore safe }$