Implement the Cubically Cube
Python 2, 476 bytes

import re
c=[[i]*9for i in range(6)]
r=lambda f:[f[a+b]for a in(0,1,2)for b in(6,3,0)]
U=lambda c:[r(c[0])]+[c[j+1][:3]+c[j or 4][3:]for j in(1,2,3,0)]+[c[5]]
y=lambda c:[r(c[0])]+c[2:5]+[c[1],r(r(r(c[5])))]
z=lambda c:[c[2],r(r(r(c[1]))),c[5],r(c[3]),c[0][::-1],c[4][::-1]]
s=input()
exec(&quot;c=&quot;+&quot;(c);c=&quot;.join(&quot;&quot;.join(&quot;zzzUz U zzUzz yyyzUzzzy zUzzz yzUzzzyyy&quot;.split()[ord(t)%11%7]*sum(map(int,n))for t,n in re.findall(&quot;([BUDLFR])(\d*)&quot;,s[:-2])))+&quot;(c)&quot;)
print sum(c[int(s[-1])])

Try it online!
A port from my Simulate A Rubik's Cube answer. The revisit prompted me to golf that by 47 bytes.