In YD SE $$I_{max}=I_0$$ Path difference at a point in front of one of shifts is$$\ riangle x=d\\Big(\\cfrac{y}{D}\\Big)=d\\Bigg(\\cfrac{\\cfrac{d}{2}}{D}\\Bigg)=\\cfrac{d^2}{2D}$$ $$\ riangle x=\\cfrac{d^2}{2(10d)}=\\cfrac{d}{20}=\\cfrac{5\\lambda}{20}=\\cfrac{\\lambda}{4}$$ Path difference is$$\\phi =\\cfrac{2\\pi}{\\lambda }=(\ riangle x)=\\cfrac{2\\pi}{\\lambda}\\Big(\\cfrac{\\lambda}{4}\\Big)$$ $$\\phi =\\cfrac{\\pi}{2}$$ So intensity at that pt is$$I=I_{max}cos^2\\Big(\\cfrac{\ heta}{2}\\Big)$$ $$I=I_0cos^2\\Big(\\cfrac{\\pi}{4}\\Big)=\\cfrac{I_0}{2}$$