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# J, 39 char

**Solution:**
```
(a.{~ a.((6|.8{.1)+.(_8&{.))&.#:"0@i.])
```

**Test example:**
```
   (a.{~ a.((6|.8{.1)+.(_8&{.))&.#:"0@i.]) '([({Enclosed})])A@B.c'
({({enclosed})})a`b.c
```

**How it works:**
```
6|.8{.1 creates binary 0 0 1 0 0 0 0 0
_8&{. ensures that binari from ascii is 8 bit
(6|.8{.1)+.(_8&{.) makes third bit 1
a.((6|.8{.1)+.(_8&{.))&.#:"0@i. searches for text on right in character vector a.,...
   ...then converts to binary, then ensures third bit is 1, then, using &. reverses the binary
a.{~ converts the reversed binary, which is an ascii number, back to character text
```